Problem: $h(t)=(t-4)(t+8)$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Answer: $\begin{aligned} (t-4)(t+8)&=0 \\\\ t-4=0&\text{ or }t+8=0 \\\\ t={4}&\text{ or }t={-8} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({4})+({-8})}{2} \\\\ &={-2} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h({-2})$ : $\begin{aligned} h({-2})&=({-2}-4)({-2}+8) \\\\ &=(-6)(6) \\\\ &=-36 \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=-8 \\\\ \text{larger }t&=4 \end{aligned}$ The vertex of the parabola is at $(-2,-36)$